Let $n\geq2$ be a positive integer and let $A,B\in M_{2}(\mathbb{C})$ be two matrices such that $AB\neq BA$ and $(AB)^n=(BA)^n$. Prove that $(AB)^n=\alpha I_2$ for some complex number $\alpha$.
I have tried to use Cayley Hamilton Theorem. Please give a hint.
Let $f\in\mathbb{C}[x]$ be the characteristic polynomial of matrices $AB$ and $BA$. Now dividing the polynomial $x^n$ by $f$ (by using division algorithm) we get that there exist $Q\in\mathbb{C}[x]$ and $a,b\in\mathbb{C}$ such that $x^n=f(x)Q(x)+ax+b$. Since it is an identity in $x$, replacing $x$ by $AB$ and $BA$ we get that $(AB)^n=aAB+bI_2$ and $(BA)^n=aBA+bI_2$. Since $(AB)^n=(BA)^n$ and $AB\neq BA$, we get that $a$ must be equal to zero and this implies that $(AB)^n=(BA)^n=bI_2$ .